In reality, you can’t have contact without friction, whether you’re taking advantage of it (fastening, braking, heating) or you need to apply lubrication to minimise its effects (bearings, sliding surfaces), it needs to be taken into account. In the virtual world however, friction adds another level of complexity, quite often requiring highly robust solver capability for what might seem like the simplest of problems. Indeed, good practice dictates that you leave it to the end of your model building and testing process because of this.
Abaqus has gained a reputation for solving difficult contact problems in part because of its ability to simulate contact with friction using features such as the unsymmetric solver, as well as the different friction models available (including a user defined model).
And so to our third in a series of contact tutorials for Abaqus which focuses on defining friction within the contact definition. This exercise will show you how to include friction as an interaction property and how to investigate its effects on a disk brake system.
In this example, the model will already be pre-defined in terms of geometry, mesh and material properties. The disk brake will have an initial rotational velocity and will make contact with the disk pad. Your only task will be to define the interactions with the two approaches provided by Abaqus, run the analysis and post-process the data. The results should show an increase in disk temperature due to heat generated through energy dissipation and the reduction of rotational velocity as a consequence of friction.
The full tutorial is available here to download: